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The following problem(p.668, 7) is from Integrals and Series [ Интегралы и ряды, А.П. Прудников, Ю.А. Брычков, О.И. Маричев.] states that


How one can show that?

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Hint: re-write as $\frac{1}{3}\int_0^1 \sum x^k dx-2\int_0^1 \sum x^{2k}dx+\frac{8}{3} \int_0^1 \sum x^{4k}\,dx$. Can you see what to do next? – L. F. Dec 10 '13 at 12:05
See the answers on this related post. – Lucian Dec 10 '13 at 12:26
I'd say the title would be more commonly translated as "Integrals and Series" – DonAntonio Dec 10 '13 at 13:55
@DonAntonio: You are right! :) – Salech Alhasov Dec 10 '13 at 13:57
up vote 4 down vote accepted

\begin{align} \sum_{k=0}^{\infty}\frac{1}{(k+1)(2k+1)(4k+1)}&=-\sum_{k=0}^\infty \frac{1}{k+1/2}+\frac{2}{3}\sum_{k=0}^{\infty} \frac{1}{k+1/4}+\frac{1}{3} \sum_{k=0}^\infty \frac{1}{k+1} \\ &=\frac{1}{3}\sum_{k=0}^\infty \frac{1}{k+1}-\frac{1}{k+1/2}+\frac{2}{3} \sum_{k=0}^{\infty} \frac{1}{k+1/4}-\frac{1}{k+1/2} \\ &=-\frac{1}{6}\sum_{k=0}^{\infty} \frac{1}{(k+1)(k+1/2)}+\frac{1}{6}\sum_{k=0}^\infty \frac{1}{(k+1/4)(k+1/2)} \\ \text{Using Gauss's Digamma Theorem, }\\ &=-\frac{1}{6}\cdot \frac{\psi(1)-\psi(1/2)}{1/3-1/2}+ \frac{1}{6}\cdot \frac{\psi(1/4)-\psi(1/2)}{1/4-1/2} \\ &=-\frac{1}{6}\cdot \log 16+\frac{1}{6}\cdot 2(\pi+\log4)=\frac{\pi}{3} \end{align}

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